# Ptolemy's Theorem

**Ptolemy's theorem** states the relationship between the diagonals and the sides of a cyclic quadrilateral. It is a powerful tool to apply to problems about inscribed quadrilaterals.

Ptolemy's TheoremIf a quadrilateral is inscribable in a circle, then the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of the opposite sides:

$AC\cdot BD = AB\cdot CD + AD\cdot BC.$ $\hspace{1.5cm}$

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## Proof

Let's prove this theorem.

Let $ABCD$ be a random quadrilateral inscribed in a circle.

The proposition will be proved if $AC\cdot BD = AB\cdot CD + AD\cdot BC.$

It's easy to see in the inscribed angles that $\angle ABD = \angle ACD, \angle BDA= \angle BCA,$ and $\angle BAC = \angle BDC.$

Let $E$ be a point on $AC$ such that $\angle EBC = \angle ABD = \angle ACD,$ then since $\angle EBC = \angle ABD$ and $\angle BCA= \angle BDA,$

$\triangle EBC \approx \triangle ABD \Leftrightarrow \dfrac{CB}{DB} = \dfrac{CE}{AD} \Leftrightarrow AD\cdot CB = DB\cdot CE. \qquad (1)$

Note that $\angle ABD = \angle EBC \Leftrightarrow \angle ABD + \angle KBE = \angle EBC + \angle KBE \Rightarrow \angle ABE = \angle CBK.$ Then since $\angle ABE= \angle CBK$ and $\angle CAB= \angle CDB,$

$\triangle ABE \approx \triangle BDC \Leftrightarrow \dfrac{AB}{DB} = \dfrac{AE}{CD} \Leftrightarrow CD\cdot AB = DB\cdot AE. \qquad (2)$

Therefore, from $(1)$ and $(2),$ we have

$\begin{aligned} AB \cdot CD + AD\cdot BC & = CE\cdot DB + AE\cdot DB \\ & = (CE+AE)DB \\ & = CA\cdot DB. \end{aligned}$

Hence proved. $_\square$

## Applications

We can prove the Pythagorean theorem using Ptolemy's theorem:

Prove that in any right-angled triangle $\triangle ABC$ where $\angle A = 90^\circ,$ $AB^2 + AC^2 = BC^2.$

Let $ABDC$ be a random rectangle inscribed in a circle.$$

$$

Applying Ptolemy's theorem in the rectangle, we get

$AD\cdot BC = AB\cdot DC + AC\cdot DB.$

But $AD= BC, AB = DC, AC = DB$ since $ABDC$ is a rectangle. Therefore,

$BC^2 = AB^2 + AC^2. \ _\square$

Once upon a time, Ptolemy let his pupil draw an equilateral triangle $ABC$ inscribed in a circle before the great mathematician depicted point $D$ and joined the red lines with other vertices, as shown below.

**Ptolemy**: Dost thou see that all the red lines have the lengths in whole integers?

**Pupil**: Indeed, master! Such an extraordinary point!

**Ptolemy**: Now if the equilateral triangle has a side length of 13, what is the sum of the three red lengths combined?

$ABCD$ is a cyclic quadrilateral with $\displaystyle \overline{AB}=11$ and $\displaystyle \overline{CD}=19$. $P$ and $Q$ are points on $\overline{AB}$ and $\overline{CD}$, respectively, such that $\displaystyle \overline{AP}=6$, $\displaystyle \overline{DQ}=7$, and $\displaystyle \overline{PQ}=27.$ Determine the length of the line segment formed when $\displaystyle \overline{PQ}$ is extended from both sides until it reaches the circle.

**Note**: The image is not drawn to scale.

## Converse

In a quadrilateral, if the product of its diagonals is equal to the sum of the products of the pairs of the opposite sides, then the quadrilateral is

inscribable.

## Ptolemy's Inequality

Given a quadrilateral $ABCD$,

$AC\cdot BD \leq AB\cdot CD + AD\cdot BC,$

where equality occurs if and only if $ABCD$ is inscribable. $_\square$

Let $I$ be a point inside quadrilateral $ABCD$ such that $\angle ABD = \angle IBC$ and $\angle ADB = \angle ICB$.

Triangle $ABD$ is similar to triangle $IBC$, so $\frac{AB}{IB}=\frac{BD}{BC}=\frac{AD}{IC} \implies AD \cdot BC = BD \cdot IC$ and $\frac{AB}{BD}=\frac{IB}{BC}$.

This gives us another pair of similar triangles: $ABI$ and $DBC$ $\implies \frac{AI}{DC}=\frac{AB}{BD} \implies AB \cdot CD = AI \cdot BD$.

Adding the two equations together,

$AB\cdot CD + AD\cdot BC = BD \cdot (IA + IC) \geq BD \cdot AC.$

The equality occurs when $I$ lies on $AC$, which means $ABCD$ is inscribable. $_\square$

Another proof requires a basic understanding of properties of inversions, especially those relevant to distance.

Consider a circle of radius 1 centred at $A$. Let $B', C',$ and $D'$ be the resultant of inverting points $B, C,$ and $D$ about this circle, respectively. Using the distance properties of inversion, we have

$\begin{aligned} AB &= \frac{1}{AB'}\\ CD &= \frac{C'D'}{AC' \cdot AD'}\\ AD &= \frac{1}{AD'}\\ BC &= \frac{B'C'}{AB' \cdot AC'}\\ AC &= \frac{1}{AC'}\\ BD &= \frac{B'D'}{AB' \cdot AD'}. \end{aligned}$

Thus, Ptolemy's inequality becomes

$\begin{aligned} AB \cdot CD + AD \cdot BC &\geq BD \cdot AC\\ \frac{1}{AB'} \cdot \frac{C'D'}{AC' \cdot AD'} + \frac{1}{AD'} \cdot \frac{B'C'}{AB' \cdot AC'} &\geq \frac{1}{AC'} \cdot \frac{B'D'}{AB' \cdot AD'}\\\\ C'D' + B'C' &\geq B'D', \end{aligned}$

which is true by triangle inequality. Therefore, Ptolemy's inequality is true. Thus proven. $_\square$

**Cite as:**Ptolemy's Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/ptolemys-theorem/